Determine an instantaneous rate of change equation given this information

Question

I'm in Grade 12 Advanced Functions and I've been stumped by a thinking question. Here it is.

Rael is investigating the rate of change of the function $y = \cos x$ on the interval $x \in [0, 2\pi]$. He notices that the graph of $y = \cos x$ passes through the $x$-axis at $45$ degrees. He also determines the instantaneous rate of change at $x = 0, \pi$, and $2\pi$ by inspection. Based on this information, determine an equation $r(x)$ to predict the instantaneous rate of change of the function $y = \cos x$ on the interval $x \in [0, 2\pi]$. Then, use your equation to calculate the exact instantaneous rate of change at $x = \pi/4$.

I know the instantaneous rate of change formula is $$\frac{f(b)-f(a)}{b-a},$$ so I don't particularly see how I'm to construct my own such equation, especially implementing this information. Any help?

Answer 1

There isn’t any “information” given by the question, so you have to find the equation for $\cos’ x$ (the derivative of $\cos x$). Usually, the following formulas are used.

$$f’(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim_{b \to x}\frac{f(b)-f(x)}{b-x}$$

Which are both equivalent (if you let $h = b-x$).

If you know about derivatives, you’ll know that $\cos’x = -\sin x$, which gives the instantaneous rate of change. If not, here’s one way to derive it.

$$\lim_{h \to 0}\frac{\cos(x+h)-\cos x}{h}$$

$$= \lim_{h \to 0}\frac{\color{blue}{\cos x\cos h}-\sin x\sin h\color{blue}{-\cos x}}{h}$$

$$= \lim_{h \to 0}\frac{\color{blue}{\cos x(\cos h-1)}-\sin x\sin h}{h}$$

$$= \lim_{h \to 0}\frac{\cos x(\cos h-1)}{h}-\lim_{h \to 0}\frac{\sin x\sin h}{h}$$

$$= \cos x\cdot\lim_{h \to 0}\frac{\cos h-1}{h}-\sin x\cdot\lim_{h \to 0}\frac{\sin h}{h}$$

Using $\lim_\limits{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_\limits{h \to 0}\frac{\cos h-1}{h} = 0$, you get

$$= \cos x\cdot 0 - \sin x\cdot 1 = \color{purple}{-\sin x}$$

The instantaneous slope/slope of the tangent at point $x_1$ can be found by $$m = f’(x_1) \implies m = -\sin(x_1)$$

The given point is $x_1 = \frac{\pi}{4}$, so you get

$$m = -\sin\bigg(\frac{\pi}{4}\bigg) = -\frac{\sqrt 2}{2}$$

Answer 2

Not sure how much calculus do you know but here is why differentiating cosine gives you negative of sine.

\begin{align} \lim_{b \to a} \frac{\cos(b)-\cos(a)}{b-a} &=\lim_{b \to a} \frac{-2 \sin\left( \frac{a+b}2\right) \sin\left(\frac{b-a}2 \right)}{b-a} \\ &=\lim_{b \to a}- \sin\left( \frac{a+b}2\right)\lim_{b \to a} \frac{ \sin\left(\frac{b-a}2 \right)}{\frac{b-a}2} \\ &=\lim_{b \to a}- \sin\left( \frac{a+b}2\right)\\ &= - \sin (a) \end{align}